Question

Check my work mode : This shows what is correct or incorrect for the work you have completed so far. It does not indicate The 38 kg uniform disk A and the bar BC are at rest and the 9-kg uniform disk D has an initial angular velocity wy of magnitude 440 rpm when the compressed spring is released and disk D contacts disk A. The system rotates freely about the vertical spindle BE. After a period of slippage, disk Drolls without slipping. Knowing that the magnitude of the final angular velocity of disk Dis 176 rpm, determine the final angular velocities of bar BC and disk A. Neglect the mass of bar BC. B 300 mm 115 mm E 00 34.74 $ rpm 2 The angular velocity of bar BC is WBC The angular velocity of disk A is WA 19.36 rpm. < Prev 7 of 9 Next > ***

Answer 1

WA o А s 115 mm mo Streure 3 = 18.431 rad's 60 Griven data: MA 38 kg = q kcy W = 440 rpm = 400 x 2T 60 = 46.077 radis, W q = 1760pm = 176x 2 T n, y care Contact points on va disks una-Wara Vy Vy = r tworo Vy=va get Vo=%0 worAWA Apply impllls & momenton syslân y DL we how more GO Po + TAPA I WA Now nie calculate mo otw. In a W, ID & moro (rotoA)-IA WA Ig (W, W2)= mouoldat rx) - IA WA ☺ know -> 0.5 x 9 x (115416332 -0.059513 kg. ml. s IA- mir? 0.5x 38 x(3007103) 1,71 ef. VD = (1.5x10} X 18.431)-( 300 100x10'y = 2.12 -0.JWA mono? mi2 2 I a D 2 mo2 2 kg.ml wn

12 all these in ean- -) 0.059513 (46077- 18.431) qvo (0.3+0.115) - (1.71 WA) -) 1.64.53 = 9(2.12-0.300.3+ 0.115)-1.71 WA 54.6453 (19.08-2.70A).CO.815)-1.7IWA 2) 1.6953 7.9182 =-12.780.415 +1.71) WA WA = 2.2162 rpm = = 21.163 rpm. . 2,2162 x 60 2TI A rut A in Vo=2-12 -0.3 WA=1.455 mils. S vsc vo 1.455 a 3.506 rad's (1 Vatro 0.3 +0.115 3.50 6x Go 2T Be = 33.481 opmf The angular velocity of bar BC [33.481 rpm. WBC The angular Velocity of dish is Az 21.163 spin.